message stringlengths 2 44.5k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 42 109k | cluster float64 5 5 | __index_level_0__ int64 84 217k |
|---|---|---|---|---|---|
Provide a correct Python 3 solution for this coding contest problem.
Find the minimum prime number greater than or equal to X.
Constraints
* 2 \le X \le 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
X
Output
Print the minimum prime number greater than or equal to X.
Examples
Input
20
Output
23
Input
2
Output
2
Input
99992
Output
100003 | instruction | 0 | 100,114 | 5 | 200,228 |
"Correct Solution:
```
import sys
input = sys.stdin.readline
X = int(input())
Num = X
while any(Num%i==0 for i in range(2,Num)):
Num += 1
print(Num)
``` | output | 1 | 100,114 | 5 | 200,229 |
Provide a correct Python 3 solution for this coding contest problem.
Find the minimum prime number greater than or equal to X.
Constraints
* 2 \le X \le 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
X
Output
Print the minimum prime number greater than or equal to X.
Examples
Input
20
Output
23
Input
2
Output
2
Input
99992
Output
100003 | instruction | 0 | 100,115 | 5 | 200,230 |
"Correct Solution:
```
x = int(input())-1
a=0
while a == 0:
a=1
x+=1
for i in range(2, x):
if x % i ==0:
a=0
break
print(x)
``` | output | 1 | 100,115 | 5 | 200,231 |
Provide a correct Python 3 solution for this coding contest problem.
Find the minimum prime number greater than or equal to X.
Constraints
* 2 \le X \le 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
X
Output
Print the minimum prime number greater than or equal to X.
Examples
Input
20
Output
23
Input
2
Output
2
Input
99992
Output
100003 | instruction | 0 | 100,116 | 5 | 200,232 |
"Correct Solution:
```
x = int(input())
while True:
for i in range(2,int(x**0.5)+1):
if x%i==0:
x += 1
break
else:
print(x)
break
``` | output | 1 | 100,116 | 5 | 200,233 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Find the minimum prime number greater than or equal to X.
Constraints
* 2 \le X \le 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
X
Output
Print the minimum prime number greater than or equal to X.
Examples
Input
20
Output
23
Input
2
Output
2
Input
99992
Output
100003
Submitted Solution:
```
X = int(input().strip())
res = X
while True:
for i in range(2, res):
if res % i == 0:
break
else:
break
res += 1
print(res)
``` | instruction | 0 | 100,117 | 5 | 200,234 |
Yes | output | 1 | 100,117 | 5 | 200,235 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Find the minimum prime number greater than or equal to X.
Constraints
* 2 \le X \le 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
X
Output
Print the minimum prime number greater than or equal to X.
Examples
Input
20
Output
23
Input
2
Output
2
Input
99992
Output
100003
Submitted Solution:
```
X=int(input())
b=X**0.5
while any(X%i==0 for i in range(2,int(b)+1)):
X+=1
else:
print(X)
``` | instruction | 0 | 100,118 | 5 | 200,236 |
Yes | output | 1 | 100,118 | 5 | 200,237 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Find the minimum prime number greater than or equal to X.
Constraints
* 2 \le X \le 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
X
Output
Print the minimum prime number greater than or equal to X.
Examples
Input
20
Output
23
Input
2
Output
2
Input
99992
Output
100003
Submitted Solution:
```
X = int(input())
p = X
while True:
for j in range(2, int(p ** 0.5) + 1):
if p % j == 0: break
else:
break
p += 1
print(p)
``` | instruction | 0 | 100,119 | 5 | 200,238 |
Yes | output | 1 | 100,119 | 5 | 200,239 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Find the minimum prime number greater than or equal to X.
Constraints
* 2 \le X \le 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
X
Output
Print the minimum prime number greater than or equal to X.
Examples
Input
20
Output
23
Input
2
Output
2
Input
99992
Output
100003
Submitted Solution:
```
X=int(input())
for i in range(X,X*2):
ok=0
for j in range(2,i):
if i%j == 0:
ok += 1
if ok == 0:
print(i)
break
``` | instruction | 0 | 100,120 | 5 | 200,240 |
Yes | output | 1 | 100,120 | 5 | 200,241 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Find the minimum prime number greater than or equal to X.
Constraints
* 2 \le X \le 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
X
Output
Print the minimum prime number greater than or equal to X.
Examples
Input
20
Output
23
Input
2
Output
2
Input
99992
Output
100003
Submitted Solution:
```
n=int(input())
prime=0
tmp = n
while prime == 0:
flag=True
for i in range(2,int(n**0.5)+1):
flag=False
break
if frag:
prime=tmp
tmp+=1
print(prime)
``` | instruction | 0 | 100,121 | 5 | 200,242 |
No | output | 1 | 100,121 | 5 | 200,243 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Find the minimum prime number greater than or equal to X.
Constraints
* 2 \le X \le 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
X
Output
Print the minimum prime number greater than or equal to X.
Examples
Input
20
Output
23
Input
2
Output
2
Input
99992
Output
100003
Submitted Solution:
```
def primelist(n):
is_prime = [True] * (n + 1)
is_prime[0] = False
is_prime[1] = False
for i in range(2, int(n**0.5) + 1):
if not is_prime[i]:
continue
for j in range(i * 2, n + 1, i):
is_prime[j] = False
return [i for i in range(n + 1) if is_prime[i]]
x = int(input())
list = primelist(x*10)
print(min([i for i in list if i>x])
``` | instruction | 0 | 100,122 | 5 | 200,244 |
No | output | 1 | 100,122 | 5 | 200,245 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Find the minimum prime number greater than or equal to X.
Constraints
* 2 \le X \le 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
X
Output
Print the minimum prime number greater than or equal to X.
Examples
Input
20
Output
23
Input
2
Output
2
Input
99992
Output
100003
Submitted Solution:
```
import sys
import math
import itertools
rl = sys.stdin.readline
x = int(rl())
while True:
if x % 2:
for i in range(3, 7//2, 2):
if x % i == 0
break
else:
print(x)
break
x += 1
``` | instruction | 0 | 100,123 | 5 | 200,246 |
No | output | 1 | 100,123 | 5 | 200,247 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Find the minimum prime number greater than or equal to X.
Constraints
* 2 \le X \le 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
X
Output
Print the minimum prime number greater than or equal to X.
Examples
Input
20
Output
23
Input
2
Output
2
Input
99992
Output
100003
Submitted Solution:
```
import math
X=int(input())
if X==2:
print(2)
else:
n=1
while n==1:
for j in range(2,int(math.sqrt(X))+1):
if j==int(math.sqrt(X)) and X%j!=0:
n=0
print(X)
elif X%j==0:
X+=1
exit()
``` | instruction | 0 | 100,124 | 5 | 200,248 |
No | output | 1 | 100,124 | 5 | 200,249 |
Provide a correct Python 3 solution for this coding contest problem.
We have a sequence of N integers: A_1, A_2, \cdots, A_N.
You can perform the following operation between 0 and K times (inclusive):
* Choose two integers i and j such that i \neq j, each between 1 and N (inclusive). Add 1 to A_i and -1 to A_j, possibly producing a negative element.
Compute the maximum possible positive integer that divides every element of A after the operations. Here a positive integer x divides an integer y if and only if there exists an integer z such that y = xz.
Constraints
* 2 \leq N \leq 500
* 1 \leq A_i \leq 10^6
* 0 \leq K \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N K
A_1 A_2 \cdots A_{N-1} A_{N}
Output
Print the maximum possible positive integer that divides every element of A after the operations.
Examples
Input
2 3
8 20
Output
7
Input
2 10
3 5
Output
8
Input
4 5
10 1 2 22
Output
7
Input
8 7
1 7 5 6 8 2 6 5
Output
5 | instruction | 0 | 100,126 | 5 | 200,252 |
"Correct Solution:
```
N, K = map(int, input().split())
A = list(map(int, input().split()))
Asum = sum(A)
div = set()
for i in range(1, int(Asum ** 0.5 + 0.5) + 1):
if Asum % i == 0:
div.add(i)
div.add(Asum//i)
ans = 1
for d in div:
R = [a % d for a in A]
R.sort()
r = sum(R) // d
l = N - r
need = sum(R[:l])
if need <= K:
ans = max(ans, d)
print(ans)
``` | output | 1 | 100,126 | 5 | 200,253 |
Provide a correct Python 3 solution for this coding contest problem.
We have a sequence of N integers: A_1, A_2, \cdots, A_N.
You can perform the following operation between 0 and K times (inclusive):
* Choose two integers i and j such that i \neq j, each between 1 and N (inclusive). Add 1 to A_i and -1 to A_j, possibly producing a negative element.
Compute the maximum possible positive integer that divides every element of A after the operations. Here a positive integer x divides an integer y if and only if there exists an integer z such that y = xz.
Constraints
* 2 \leq N \leq 500
* 1 \leq A_i \leq 10^6
* 0 \leq K \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N K
A_1 A_2 \cdots A_{N-1} A_{N}
Output
Print the maximum possible positive integer that divides every element of A after the operations.
Examples
Input
2 3
8 20
Output
7
Input
2 10
3 5
Output
8
Input
4 5
10 1 2 22
Output
7
Input
8 7
1 7 5 6 8 2 6 5
Output
5 | instruction | 0 | 100,127 | 5 | 200,254 |
"Correct Solution:
```
n,k,*a=map(int,open(0).read().split())
s=sum(a)
b=[]
for i in range(1,int(s**.5)+1):b+=[s//i,i]*(s%i<1)
m=1
for i in b:c=sorted(j%i for j in a);m=max(m,i*(sum(c[:-sum(c)//i])<=k))
print(m)
``` | output | 1 | 100,127 | 5 | 200,255 |
Provide a correct Python 3 solution for this coding contest problem.
We have a sequence of N integers: A_1, A_2, \cdots, A_N.
You can perform the following operation between 0 and K times (inclusive):
* Choose two integers i and j such that i \neq j, each between 1 and N (inclusive). Add 1 to A_i and -1 to A_j, possibly producing a negative element.
Compute the maximum possible positive integer that divides every element of A after the operations. Here a positive integer x divides an integer y if and only if there exists an integer z such that y = xz.
Constraints
* 2 \leq N \leq 500
* 1 \leq A_i \leq 10^6
* 0 \leq K \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N K
A_1 A_2 \cdots A_{N-1} A_{N}
Output
Print the maximum possible positive integer that divides every element of A after the operations.
Examples
Input
2 3
8 20
Output
7
Input
2 10
3 5
Output
8
Input
4 5
10 1 2 22
Output
7
Input
8 7
1 7 5 6 8 2 6 5
Output
5 | instruction | 0 | 100,128 | 5 | 200,256 |
"Correct Solution:
```
n,k=map(int,input().split())
a=list(map(int,input().split()))
m=sum(a)
cd=set(())
for i in range(1,int(m**0.5)+2):
if m%i==0:
cd.add(i)
cd.add(m//i)
cd=list(cd)
cd.sort(reverse=True)
def func(x):
r=[ai%x for ai in a]
r.sort()
tmp=0
sr=[0]
for ri in r:
tmp+=ri
sr.append(tmp)
for i in range(n+1):
tmp0=sr[i]
tmp1=(n-i)*x-(sr[-1]-sr[i])
if tmp0==tmp1 and tmp0<=k:
return True
return False
for x in cd:
if x==1:
print(1)
exit()
if func(x):
print(x)
exit()
``` | output | 1 | 100,128 | 5 | 200,257 |
Provide a correct Python 3 solution for this coding contest problem.
We have a sequence of N integers: A_1, A_2, \cdots, A_N.
You can perform the following operation between 0 and K times (inclusive):
* Choose two integers i and j such that i \neq j, each between 1 and N (inclusive). Add 1 to A_i and -1 to A_j, possibly producing a negative element.
Compute the maximum possible positive integer that divides every element of A after the operations. Here a positive integer x divides an integer y if and only if there exists an integer z such that y = xz.
Constraints
* 2 \leq N \leq 500
* 1 \leq A_i \leq 10^6
* 0 \leq K \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N K
A_1 A_2 \cdots A_{N-1} A_{N}
Output
Print the maximum possible positive integer that divides every element of A after the operations.
Examples
Input
2 3
8 20
Output
7
Input
2 10
3 5
Output
8
Input
4 5
10 1 2 22
Output
7
Input
8 7
1 7 5 6 8 2 6 5
Output
5 | instruction | 0 | 100,131 | 5 | 200,262 |
"Correct Solution:
```
n,k = list(map(int,input().split()))
a = list(map(int,input().split()))
candidate=set()
result=0
x=1
while x**2<=(sum(a)+1):
if sum(a)%x==0:
candidate.add(x)
candidate.add(int(sum(a)/x))
x+=1
# print(candidate)
for i in candidate:
b=[0]*len(a)
for j in range(len(a)):
b[j]=int(a[j]%i)
b.sort()
if sum(b[:int(len(b)-sum(b)/i)])<=k:
result=max(result,i)
# need=0
# need+=sum(b)/i
# if need<=k:
print(result)
``` | output | 1 | 100,131 | 5 | 200,263 |
Provide a correct Python 3 solution for this coding contest problem.
We have a sequence of N integers: A_1, A_2, \cdots, A_N.
You can perform the following operation between 0 and K times (inclusive):
* Choose two integers i and j such that i \neq j, each between 1 and N (inclusive). Add 1 to A_i and -1 to A_j, possibly producing a negative element.
Compute the maximum possible positive integer that divides every element of A after the operations. Here a positive integer x divides an integer y if and only if there exists an integer z such that y = xz.
Constraints
* 2 \leq N \leq 500
* 1 \leq A_i \leq 10^6
* 0 \leq K \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N K
A_1 A_2 \cdots A_{N-1} A_{N}
Output
Print the maximum possible positive integer that divides every element of A after the operations.
Examples
Input
2 3
8 20
Output
7
Input
2 10
3 5
Output
8
Input
4 5
10 1 2 22
Output
7
Input
8 7
1 7 5 6 8 2 6 5
Output
5 | instruction | 0 | 100,132 | 5 | 200,264 |
"Correct Solution:
```
n,k = map(int, input().split())
a = list(map(int, input().split()))
t = sum(a)
r = set()
r.add(t)
for i in range(2,int(t**0.5)+2):
if t%i == 0:
r.add(i)
r.add(t//i)
ans = 1
for i in r:
f = [x%i for x in a]
f.sort()
if sum(f[:-sum(f)//i]) <= k:
ans = max(ans,i)
print(ans)
``` | output | 1 | 100,132 | 5 | 200,265 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a sequence of N integers: A_1, A_2, \cdots, A_N.
You can perform the following operation between 0 and K times (inclusive):
* Choose two integers i and j such that i \neq j, each between 1 and N (inclusive). Add 1 to A_i and -1 to A_j, possibly producing a negative element.
Compute the maximum possible positive integer that divides every element of A after the operations. Here a positive integer x divides an integer y if and only if there exists an integer z such that y = xz.
Constraints
* 2 \leq N \leq 500
* 1 \leq A_i \leq 10^6
* 0 \leq K \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N K
A_1 A_2 \cdots A_{N-1} A_{N}
Output
Print the maximum possible positive integer that divides every element of A after the operations.
Examples
Input
2 3
8 20
Output
7
Input
2 10
3 5
Output
8
Input
4 5
10 1 2 22
Output
7
Input
8 7
1 7 5 6 8 2 6 5
Output
5
Submitted Solution:
```
N, K = map(int, input().split())
A = list(map(int, input().split()))
divs = []
maxA = sum(A)
for i in range(1, int(maxA ** 0.5) + 1):
if maxA % i == 0:
divs.append(i)
divs.append(maxA // i)
divs.sort(reverse=True)
for d in divs:
rest = [a % d for a in A]
rest.sort(reverse=True)
restSum = sum(rest) // d
cnt = 0
for i in range(restSum):
cnt += d - rest[i]
if cnt <= K:
print(d)
exit()
``` | instruction | 0 | 100,135 | 5 | 200,270 |
Yes | output | 1 | 100,135 | 5 | 200,271 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a sequence of N integers: A_1, A_2, \cdots, A_N.
You can perform the following operation between 0 and K times (inclusive):
* Choose two integers i and j such that i \neq j, each between 1 and N (inclusive). Add 1 to A_i and -1 to A_j, possibly producing a negative element.
Compute the maximum possible positive integer that divides every element of A after the operations. Here a positive integer x divides an integer y if and only if there exists an integer z such that y = xz.
Constraints
* 2 \leq N \leq 500
* 1 \leq A_i \leq 10^6
* 0 \leq K \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N K
A_1 A_2 \cdots A_{N-1} A_{N}
Output
Print the maximum possible positive integer that divides every element of A after the operations.
Examples
Input
2 3
8 20
Output
7
Input
2 10
3 5
Output
8
Input
4 5
10 1 2 22
Output
7
Input
8 7
1 7 5 6 8 2 6 5
Output
5
Submitted Solution:
```
n,k=map(int,input().split())
a=list(map(int,input().split()))
a_sum=sum(a)
def devisor(x):
lst=[]
i=1
y=x
while i**2<=y:
if x%i==0:
lst.append(i)
if i!=x//i:
lst.append(x//i)
i+=1
lst.sort()
return lst
re=devisor(a_sum)
re.sort(reverse=True)
for u in re:
w=[]
for v in a:
w.append(v%u)
w.sort()
b=sum(w)
count=0
for i in range(b//u):
w[-i-1]-=u
for i in w:
if i>=0:
count+=i
else:
break
if count<=k:
print(u)
break
``` | instruction | 0 | 100,136 | 5 | 200,272 |
Yes | output | 1 | 100,136 | 5 | 200,273 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a sequence of N integers: A_1, A_2, \cdots, A_N.
You can perform the following operation between 0 and K times (inclusive):
* Choose two integers i and j such that i \neq j, each between 1 and N (inclusive). Add 1 to A_i and -1 to A_j, possibly producing a negative element.
Compute the maximum possible positive integer that divides every element of A after the operations. Here a positive integer x divides an integer y if and only if there exists an integer z such that y = xz.
Constraints
* 2 \leq N \leq 500
* 1 \leq A_i \leq 10^6
* 0 \leq K \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N K
A_1 A_2 \cdots A_{N-1} A_{N}
Output
Print the maximum possible positive integer that divides every element of A after the operations.
Examples
Input
2 3
8 20
Output
7
Input
2 10
3 5
Output
8
Input
4 5
10 1 2 22
Output
7
Input
8 7
1 7 5 6 8 2 6 5
Output
5
Submitted Solution:
```
n, k = map(int, input().split())
a = list(map(int, input().split()))
s = sum(a)
d = []
for i in range(1, s):
if s % i == 0:
d.append(i)
if s <= i * i:
break
if s % i == 0:
d.append(s // i)
d = list(reversed(sorted(d)))
for di in d:
b = []
for ai in a:
b.append(ai % di)
b = list(sorted(b))
cnt = 0
il = 0
ir = n - 1
while il < ir:
if b[il] + b[ir] == di:
cnt += b[il]
b[il] = 0
b[ir] = di
il += 1
ir -= 1
elif b[il] < di - b[ir]:
cnt += b[il]
b[il] = 0
b[ir] += b[il]
il += 1
else:
cnt += di - b[ir]
b[ir] = di
b[il] -= di - b[ir]
ir -= 1
if cnt <= k:
print(di)
break
``` | instruction | 0 | 100,138 | 5 | 200,276 |
No | output | 1 | 100,138 | 5 | 200,277 |
Provide a correct Python 3 solution for this coding contest problem.
Nukes has an integer that can be represented as the bitwise OR of one or more integers between A and B (inclusive). How many possible candidates of the value of Nukes's integer there are?
Constraints
* 1 ≤ A ≤ B < 2^{60}
* A and B are integers.
Input
The input is given from Standard Input in the following format:
A
B
Output
Print the number of possible candidates of the value of Nukes's integer.
Examples
Input
7
9
Output
4
Input
65
98
Output
63
Input
271828182845904523
314159265358979323
Output
68833183630578410 | instruction | 0 | 100,193 | 5 | 200,386 |
"Correct Solution:
```
import sys
def solve():
a = int(input())
b = int(input())
if a == b:
print(1)
return
t = a ^ b
N = bitlen(t)
a = a & (2**N - 1)
b = b & (2**N - 1)
blen = bitlen(b)
sb = b & (2**(blen - 1) - 1)
sblen = bitlen(sb)
s = 2**sblen - 1
ymax = b | s
T = 2**(N - 1)
ans = T - a
if ymax < T + a:
ans += ymax + 1 - a
else:
ans += T
print(ans)
def bitlen(x):
return 1 + bitlen(x >> 1) if x else 0
if __name__ == '__main__':
solve()
``` | output | 1 | 100,193 | 5 | 200,387 |
Provide a correct Python 3 solution for this coding contest problem.
Nukes has an integer that can be represented as the bitwise OR of one or more integers between A and B (inclusive). How many possible candidates of the value of Nukes's integer there are?
Constraints
* 1 ≤ A ≤ B < 2^{60}
* A and B are integers.
Input
The input is given from Standard Input in the following format:
A
B
Output
Print the number of possible candidates of the value of Nukes's integer.
Examples
Input
7
9
Output
4
Input
65
98
Output
63
Input
271828182845904523
314159265358979323
Output
68833183630578410 | instruction | 0 | 100,194 | 5 | 200,388 |
"Correct Solution:
```
import sys
readline = sys.stdin.readline
def calc(A, B):
if A == B:
return 1
for i in range(61, -1, -1):
if A&(1<<i) and B&(1<<i):
A ^= 1<<i
B ^= 1<<i
if not A&(1<<i) and B&(1<<i):
break
Bd = B ^ (1<<(B.bit_length())-1)
BM = ((1<<Bd.bit_length())-1)
B = B | BM
C = (1<<(B.bit_length())-1)
res = (B-A+1) + (C-max(BM+1, A))
return res
A = int(readline())
B = int(readline())
print(calc(A, B))
``` | output | 1 | 100,194 | 5 | 200,389 |
Provide a correct Python 3 solution for this coding contest problem.
Nukes has an integer that can be represented as the bitwise OR of one or more integers between A and B (inclusive). How many possible candidates of the value of Nukes's integer there are?
Constraints
* 1 ≤ A ≤ B < 2^{60}
* A and B are integers.
Input
The input is given from Standard Input in the following format:
A
B
Output
Print the number of possible candidates of the value of Nukes's integer.
Examples
Input
7
9
Output
4
Input
65
98
Output
63
Input
271828182845904523
314159265358979323
Output
68833183630578410 | instruction | 0 | 100,195 | 5 | 200,390 |
"Correct Solution:
```
import math
A = int(input())
B = int(input())
if B < A: A,B = B,A
if A == B:
print(1)
else:
t = int(math.floor(math.log2(B)))
while (A & 2**t) == (B & 2**t):
if A & 2**t:
A -= 2**t
B -= 2**t
t-=1
ans = 2**t - A # [A,2^t-1]
r = t-1
while 2**t + 2**r > B and r>=0:
r-=1
# [2^t,2^t+2^{r+1}-1] \cup [2^t+A,2^{t+1}-1]
C = 2**(r+1)
if C < A:
ans += C + 2**t - A
else:
ans += 2**t
print(ans)
``` | output | 1 | 100,195 | 5 | 200,391 |
Provide a correct Python 3 solution for this coding contest problem.
Nukes has an integer that can be represented as the bitwise OR of one or more integers between A and B (inclusive). How many possible candidates of the value of Nukes's integer there are?
Constraints
* 1 ≤ A ≤ B < 2^{60}
* A and B are integers.
Input
The input is given from Standard Input in the following format:
A
B
Output
Print the number of possible candidates of the value of Nukes's integer.
Examples
Input
7
9
Output
4
Input
65
98
Output
63
Input
271828182845904523
314159265358979323
Output
68833183630578410 | instruction | 0 | 100,196 | 5 | 200,392 |
"Correct Solution:
```
import sys
def solve():
a = int(input())
b = int(input())
if a == b:
print(1)
return
t = a ^ b
N = len(bin(t)) - 2
t = 1 << N
a = a & (t - 1)
b = b & (t - 1)
blen = len(bin(b)) - 2
sb = b & (2**(blen - 1) - 1)
if sb == 0:
sblen = 0
else:
sblen = len(bin(sb)) - 2
s = (1<<sblen) - 1
ymax = b | s
T = 1<<(N - 1)
ans = T - a
if ymax < T + a:
ans += ymax - T + 1 + T - a
else:
ans += T
print(ans)
if __name__ == '__main__':
solve()
``` | output | 1 | 100,196 | 5 | 200,393 |
Provide a correct Python 3 solution for this coding contest problem.
Nukes has an integer that can be represented as the bitwise OR of one or more integers between A and B (inclusive). How many possible candidates of the value of Nukes's integer there are?
Constraints
* 1 ≤ A ≤ B < 2^{60}
* A and B are integers.
Input
The input is given from Standard Input in the following format:
A
B
Output
Print the number of possible candidates of the value of Nukes's integer.
Examples
Input
7
9
Output
4
Input
65
98
Output
63
Input
271828182845904523
314159265358979323
Output
68833183630578410 | instruction | 0 | 100,197 | 5 | 200,394 |
"Correct Solution:
```
a=int(input())
b=int(input())
if a==b:
print(1)
exit()
abin=bin(a)[2:]
bbin=bin(b)[2:]
la=len(abin)
lb=len(bbin)
if la==lb:
while abin[0]==bbin[0]:
abin=abin[1:]
bbin=bbin[1:]
while len(abin)>1 and abin[0]=="0":
abin=abin[1:]
cbin=bbin[1:]
while cbin and cbin[0]=="0":
cbin=cbin[1:]
c=2**(len(cbin))-1
a=int(abin,2)
b=int(bbin,2)
lb=len(bbin)
lbnd=2**(lb-1)+c
hbnd=2**(lb-1)+a
if lbnd>=hbnd:
print(2**lb-a)
else:
print(2**(lb-1)-a+lbnd-a+1)
``` | output | 1 | 100,197 | 5 | 200,395 |
Provide a correct Python 3 solution for this coding contest problem.
Nukes has an integer that can be represented as the bitwise OR of one or more integers between A and B (inclusive). How many possible candidates of the value of Nukes's integer there are?
Constraints
* 1 ≤ A ≤ B < 2^{60}
* A and B are integers.
Input
The input is given from Standard Input in the following format:
A
B
Output
Print the number of possible candidates of the value of Nukes's integer.
Examples
Input
7
9
Output
4
Input
65
98
Output
63
Input
271828182845904523
314159265358979323
Output
68833183630578410 | instruction | 0 | 100,198 | 5 | 200,396 |
"Correct Solution:
```
# D
res = 0
A = int(input())
B = int(input())
def pow2_floor_log2(x):
res = 1
while res*2 <= x:
res *= 2
return res
if A == B:
res += 1
while A < B:
# find log2A and log2B
C = pow2_floor_log2(B)
if A < C:
# A to C
res += C - A
if B > C:
B_ = C + pow2_floor_log2(B - C)*2 - 1
else:
B_ = C
if C + A <= B_:
res += C
break
else:
res += (C - A) + (B_ - C + 1)
break
else:
A -= C
B -= C
print(res)
``` | output | 1 | 100,198 | 5 | 200,397 |
Provide a correct Python 3 solution for this coding contest problem.
Nukes has an integer that can be represented as the bitwise OR of one or more integers between A and B (inclusive). How many possible candidates of the value of Nukes's integer there are?
Constraints
* 1 ≤ A ≤ B < 2^{60}
* A and B are integers.
Input
The input is given from Standard Input in the following format:
A
B
Output
Print the number of possible candidates of the value of Nukes's integer.
Examples
Input
7
9
Output
4
Input
65
98
Output
63
Input
271828182845904523
314159265358979323
Output
68833183630578410 | instruction | 0 | 100,199 | 5 | 200,398 |
"Correct Solution:
```
def solve(a, b):
if a == b:
return 1
f = 1 << 60
r = 0
while f:
if r == 0 and b & f and a & f == 0:
r = f
elif r > 0 and b & f:
k = f << 1
break
f >>= 1
else:
k = 1
a &= r - 1
if k > a:
return (r << 1) - a
return 2 * (r - a) + k
a = int(input())
b = int(input())
print(solve(a, b))
``` | output | 1 | 100,199 | 5 | 200,399 |
Provide a correct Python 3 solution for this coding contest problem.
Nukes has an integer that can be represented as the bitwise OR of one or more integers between A and B (inclusive). How many possible candidates of the value of Nukes's integer there are?
Constraints
* 1 ≤ A ≤ B < 2^{60}
* A and B are integers.
Input
The input is given from Standard Input in the following format:
A
B
Output
Print the number of possible candidates of the value of Nukes's integer.
Examples
Input
7
9
Output
4
Input
65
98
Output
63
Input
271828182845904523
314159265358979323
Output
68833183630578410 | instruction | 0 | 100,200 | 5 | 200,400 |
"Correct Solution:
```
import sys
readline = sys.stdin.readline
readlines = sys.stdin.readlines
from functools import lru_cache
A = int(readline())
B = int(readline())
@lru_cache(None)
def can_make(A,B):
# [A,B] の元のorで、{B} 以外の方法でBを作れるかどうか。たいていTrue
if A == B:
return False
if B%2 == 0:
return can_make((A+1)//2,B//2)
# Bが奇数なら、{B-1,B}が満たす
return True
@lru_cache(None)
def F(A,B):
if A==B:
return 1
# 偶数の作り方
ev = F((A+1)//2,B//2)
# 奇数の作り方
if A%2==0:
A += 1
if B%2 == 1:
od = F(A//2,B//2)
else:
# 上位の桁では基本的にはA//2～B//2が使える。
# ただし、{B} 以外でないといけない
od = F(A//2,B//2)
if not can_make(A//2,B//2):
od -= 1
return ev + od
answer = F(A,B)
print(answer)
``` | output | 1 | 100,200 | 5 | 200,401 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nukes has an integer that can be represented as the bitwise OR of one or more integers between A and B (inclusive). How many possible candidates of the value of Nukes's integer there are?
Constraints
* 1 ≤ A ≤ B < 2^{60}
* A and B are integers.
Input
The input is given from Standard Input in the following format:
A
B
Output
Print the number of possible candidates of the value of Nukes's integer.
Examples
Input
7
9
Output
4
Input
65
98
Output
63
Input
271828182845904523
314159265358979323
Output
68833183630578410
Submitted Solution:
```
a,b=int(input()),int(input())
if a==b:
print(1)
else:
r=60
while (a&(1<<r))==(b&(1<<r)):
r-=1
t=(1<<(r+1))
a&=t-1
b&=t-1
k=r-1
while k>=0 and (b&(1<<k))==0:
k-=1
ans=t-a
if a>(1<<(k+1)):
ans-=a-(1<<(k+1))
print(ans)
``` | instruction | 0 | 100,201 | 5 | 200,402 |
Yes | output | 1 | 100,201 | 5 | 200,403 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nukes has an integer that can be represented as the bitwise OR of one or more integers between A and B (inclusive). How many possible candidates of the value of Nukes's integer there are?
Constraints
* 1 ≤ A ≤ B < 2^{60}
* A and B are integers.
Input
The input is given from Standard Input in the following format:
A
B
Output
Print the number of possible candidates of the value of Nukes's integer.
Examples
Input
7
9
Output
4
Input
65
98
Output
63
Input
271828182845904523
314159265358979323
Output
68833183630578410
Submitted Solution:
```
a = int(input())
b = int(input())
if a == b:
print(1)
exit()
# 二進数xの長さを求める時、len(bin(x)) - 2 では、xが0のとき1を返してしまい、バグる原因になる
# よって、二進数の長さを計算する関数を用意しておく
def bitlen(x):
return 1 + bitlen(x >> 1) if x else 0
t = a ^ b
n = bitlen(t)
T = 1 << (n - 1)
a = a & (T * 2 - 1)
b = b & (T * 2 - 1)
sb = b & (T - 1)
b_max = (2 ** bitlen(sb) - 1) | T
if (T + a) > b_max:
print((b_max - a + 1) + (2 * T - 1 - (T + a) + 1))
else:
print(2 * T - 1 - a + 1)
``` | instruction | 0 | 100,202 | 5 | 200,404 |
Yes | output | 1 | 100,202 | 5 | 200,405 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nukes has an integer that can be represented as the bitwise OR of one or more integers between A and B (inclusive). How many possible candidates of the value of Nukes's integer there are?
Constraints
* 1 ≤ A ≤ B < 2^{60}
* A and B are integers.
Input
The input is given from Standard Input in the following format:
A
B
Output
Print the number of possible candidates of the value of Nukes's integer.
Examples
Input
7
9
Output
4
Input
65
98
Output
63
Input
271828182845904523
314159265358979323
Output
68833183630578410
Submitted Solution:
```
a=int(input())
b=int(input())
if a==b:
print(1)
exit()
abin=bin(a)[2:]
bbin=bin(b)[2:]
la=len(abin)
lb=len(bbin)
if la==lb:
while abin[0]==bbin[0]:
abin=abin[1:]
bbin=bbin[1:]
while len(abin)>1 and abin[0]=="0":
abin=abin[1:]
cbin=bbin[1:]
while len(cbin)>1 and cbin[0]=="0":
cbin=cbin[1:]
c=2**(len(cbin))-1
a=int(abin,2)
b=int(bbin,2)
lb=len(bbin)
lbnd=2**(lb-1)+c
hbnd=2**(lb-1)+a
if lbnd>=hbnd:
print(2**lb-a)
else:
print(2**(lb-1)-a+lbnd-a+1)
``` | instruction | 0 | 100,203 | 5 | 200,406 |
No | output | 1 | 100,203 | 5 | 200,407 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nukes has an integer that can be represented as the bitwise OR of one or more integers between A and B (inclusive). How many possible candidates of the value of Nukes's integer there are?
Constraints
* 1 ≤ A ≤ B < 2^{60}
* A and B are integers.
Input
The input is given from Standard Input in the following format:
A
B
Output
Print the number of possible candidates of the value of Nukes's integer.
Examples
Input
7
9
Output
4
Input
65
98
Output
63
Input
271828182845904523
314159265358979323
Output
68833183630578410
Submitted Solution:
```
# D
import math
res = 0
A = int(input())
B = int(input())
while True:
# find log2A and log2B
C = 2**math.floor(math.log2(B))
if A < C:
# A to C
res += C - A
if B > C:
B_ = C + 2**math.floor(math.log2(B - C) + 1) - 1
if C + A <= B_:
res += C
break
else:
res += (C - A) + (B_ - C + 1)
break
else:
A -= C
B -= C
print(res)
``` | instruction | 0 | 100,204 | 5 | 200,408 |
No | output | 1 | 100,204 | 5 | 200,409 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nukes has an integer that can be represented as the bitwise OR of one or more integers between A and B (inclusive). How many possible candidates of the value of Nukes's integer there are?
Constraints
* 1 ≤ A ≤ B < 2^{60}
* A and B are integers.
Input
The input is given from Standard Input in the following format:
A
B
Output
Print the number of possible candidates of the value of Nukes's integer.
Examples
Input
7
9
Output
4
Input
65
98
Output
63
Input
271828182845904523
314159265358979323
Output
68833183630578410
Submitted Solution:
```
A=int(input())
B=int(input())
S=[]
times=len(format(A^B,"b"))
for i in range(A,B+1):
S+=[i]
D=S[:] #Dは基礎集合
SET=set(S)
#for t in range(times):
for i in range(len(S)):
for j in range(len(D)):
SET.add(S[i]|D[j])
print(len(SET))
``` | instruction | 0 | 100,205 | 5 | 200,410 |
No | output | 1 | 100,205 | 5 | 200,411 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nukes has an integer that can be represented as the bitwise OR of one or more integers between A and B (inclusive). How many possible candidates of the value of Nukes's integer there are?
Constraints
* 1 ≤ A ≤ B < 2^{60}
* A and B are integers.
Input
The input is given from Standard Input in the following format:
A
B
Output
Print the number of possible candidates of the value of Nukes's integer.
Examples
Input
7
9
Output
4
Input
65
98
Output
63
Input
271828182845904523
314159265358979323
Output
68833183630578410
Submitted Solution:
```
import math as mt
def diff(bita,bitb,x):
if 2 ** (length - x) < b:
print((2 ** dega - b) + 2 ** (length - x) + 2 ** dega - b)
else:
print(2 ** (dega + 1) - b)
a = int(input())
b = int(input())
if b > a:
a,b = b,a
dega = int(mt.log(a, 2))
degb = int(mt.log(b, 2))
if a==b:
print(1)
exit()
if dega != degb:
bita = "{0:b}".format(a)
bitb = "{0:b}".format(b)
length = len(bita)
for x in range(1, length + 1):
if x == length:
break
if bita[x] == '1':
break
diff(bita, bitb, x)
else:
bita = "{0:b}".format(a)
bitb = "{0:b}".format(b)
length = len(bita)
for x in range(1, length + 1):
if x == length:
break
if bita[x] == '1' and bitb[x] == '0':
break
print(2 ** dega + 2 ** (length - x) - b)
``` | instruction | 0 | 100,206 | 5 | 200,412 |
No | output | 1 | 100,206 | 5 | 200,413 |
Provide a correct Python 3 solution for this coding contest problem.
Example
Input
2 2 2 0 0 0 5
Output
1 3 3 1 0 | instruction | 0 | 100,287 | 5 | 200,574 |
"Correct Solution:
```
import sys
readline = sys.stdin.readline
write = sys.stdout.write
def solve():
X, Y, Z, A, B, C, N = map(int, readline().split())
S = [0]*max(2*(X+Y+Z+1), 2*N+2)
for k in range(N):
S[k] = k*(k+1)//2
for k in range(N, X+Y+Z+1):
S[k] = k*(k+1)//2 + S[k-N]
def calc(k, x, y, z):
return S[k] - S[k-x] - S[k-y] - S[k-z] + S[k-(x+y)] + S[k-(y+z)] + S[k-(z+x)] - S[k-(x+y+z)]
ans = [0]*N
for b in range(N):
k = (X+Y+Z - b - 1) // N * N + b
ans[b] = (
calc(k+1, X-A, Y-B, Z-C)
+ calc(k, A, Y-B, Z-C)
+ calc(k, X-A, B, Z-C)
+ calc(k, X-A, Y-B, C)
+ calc(k-1, A, B, Z-C)
+ calc(k-1, X-A, B, C)
+ calc(k-1, A, Y-B, C)
+ calc(k-2, A, B, C)
)
write(" ".join(map(str, ans)))
write("\n")
solve()
``` | output | 1 | 100,287 | 5 | 200,575 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(s, t): report the minimum element in as, as+1, ..., at.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 s t
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(s, t).
Output
For each find operation, print the minimum value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0 2
1 1 2
Output
1
2
Input
1 3
1 0 0
0 0 0 5
1 0 0
Output
2147483647
5 | instruction | 0 | 100,303 | 5 | 200,606 |
"Correct Solution:
```
import sys
readline = sys.stdin.readline
write = sys.stdout.write
n, q = map(int, input().split())
INF = 2**31-1
LV = (n-1).bit_length()
N0 = 2**LV
data = [INF]*(2*N0)
lazy = [None]*(2*N0)
def gindex(l, r):
L = (l + N0) >> 1; R = (r + N0) >> 1
lc = 0 if l & 1 else (L & -L).bit_length()
rc = 0 if r & 1 else (R & -R).bit_length()
for i in range(LV):
if rc <= i:
yield R
if L < R and lc <= i:
yield L
L >>= 1; R >>= 1
def propagates(*ids):
for i in reversed(ids):
v = lazy[i-1]
if v is None:
continue
lazy[2*i-1] = data[2*i-1] = lazy[2*i] = data[2*i] = v
lazy[i-1] = None
def update(l, r, x):
*ids, = gindex(l, r)
propagates(*ids)
L = N0 + l; R = N0 + r
while L < R:
if R & 1:
R -= 1
lazy[R-1] = data[R-1] = x
if L & 1:
lazy[L-1] = data[L-1] = x
L += 1
L >>= 1; R >>= 1
for i in ids:
data[i-1] = min(data[2*i-1], data[2*i])
def query(l, r):
propagates(*gindex(l, r))
L = N0 + l; R = N0 + r
s = INF
while L < R:
if R & 1:
R -= 1
s = min(s, data[R-1])
if L & 1:
s = min(s, data[L-1])
L += 1
L >>= 1; R >>= 1
return s
ans = []
for i in range(q):
t, *cmd = map(int, readline().split())
if t:
s, t = cmd
ans.append(str(query(s, t+1)))
else:
s, t, x = cmd
update(s, t+1, x)
write("\n".join(ans))
write("\n")
``` | output | 1 | 100,303 | 5 | 200,607 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(s, t): report the minimum element in as, as+1, ..., at.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 s t
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(s, t).
Output
For each find operation, print the minimum value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0 2
1 1 2
Output
1
2
Input
1 3
1 0 0
0 0 0 5
1 0 0
Output
2147483647
5 | instruction | 0 | 100,304 | 5 | 200,608 |
"Correct Solution:
```
import math
class SegmentTree:
__slots__ = ["rank", "elem_size", "tree_size", "tree", "lazy", "default_value"]
def __init__(self, a: list, default: int):
self.default_value = default
real_size = len(a)
self.rank = math.ceil(math.log2(real_size))
self.elem_size = 1 << self.rank
self.tree_size = 2 * self.elem_size
self.tree = [default]*self.elem_size + a + [default]*(self.elem_size - real_size)
self.lazy = [None]*self.tree_size
# self.init_tree()
def init_tree(self):
tree = self.tree
for i in range(self.elem_size-1, 0, -1):
left, right = tree[i << 1], tree[(i << 1)+1]
# ===== change me =====
tree[i] = left if left < right else right
def process_query(self, l: int, r: int, value: int = None):
'''[x, y)'''
tree, lazy, elem_size, rank = self.tree, self.lazy, self.elem_size, self.rank-1
l, r, targets, p_l, p_r, l_rank, r_rank = l+elem_size, r+elem_size, [], 0, 0, 0, 0
t_ap = targets.append
while l < r:
if l & 1:
t_ap(l)
p_l = p_l or l >> 1
l_rank = l_rank or rank
l += 1
if r & 1:
r -= 1
t_ap(r)
p_r = p_r or r >> 1
r_rank = r_rank or rank
l >>= 1
r >>= 1
rank -= 1
deepest = (p_l, p_r)
paths = [[p_l >> n for n in range(l_rank-1, -1, -1)], [p_r >> n for n in range(r_rank-1, -1, -1)]]
for a in paths:
for i in a:
if lazy[i] is None:
continue
# ===== change me =====
tree[i] = lazy[i]
if i < elem_size:
lazy[i << 1] = lazy[i]
lazy[(i << 1)+1] = lazy[i]
lazy[i] = None
result = self.default_value
for i in targets:
v = value if value is not None else lazy[i]
# ===== change me =====
if v is not None:
if i < elem_size:
lazy[i << 1] = v
lazy[(i << 1)+1] = v
tree[i] = v
lazy[i] = None
if result > tree[i]:
result = tree[i]
self.update_tree(deepest)
return result
def update_tree(self, indexes: tuple):
''' ????????????lazy?????¨????????¬????????§???????????¨???????????¨???????????? '''
tree, lazy = self.tree, self.lazy
for k in indexes:
while k > 0:
left, right = k << 1, (k << 1)+1
# ===== change me =====
l_value = tree[left] if lazy[left] is None else lazy[left]
r_value = tree[right] if lazy[right] is None else lazy[right]
tree[k] = l_value if l_value < r_value else r_value
k >>= 1
n, q = map(int, input().split())
rmq = SegmentTree([2**31-1]*n, 2**31-1)
ans = []
append = ans.append
for _ in [0]*q:
l = list(map(int, input().split()))
if l[0] == 0:
rmq.process_query(l[1], l[2]+1, l[3])
else:
a = rmq.process_query(l[1], l[2]+1)
append(a)
print("\n".join([str(n) for n in ans]))
``` | output | 1 | 100,304 | 5 | 200,609 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(s, t): report the minimum element in as, as+1, ..., at.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 s t
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(s, t).
Output
For each find operation, print the minimum value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0 2
1 1 2
Output
1
2
Input
1 3
1 0 0
0 0 0 5
1 0 0
Output
2147483647
5 | instruction | 0 | 100,305 | 5 | 200,610 |
"Correct Solution:
```
import sys
readline = sys.stdin.readline
write = sys.stdout.write
N, Q = map(int, input().split())
INF = 2**31-1
LV = (N-1).bit_length()
N0 = 2**LV
data = [INF]*(2*N0)
lazy = [None]*(2*N0)
def gindex(l, r):
L = l + N0; R = r + N0
lm = (L // (L & -L)) >> 1
rm = (R // (R & -R)) >> 1
while L < R:
if R <= rm:
yield R
if L <= lm:
yield L
L >>= 1; R >>= 1
while L:
yield L
L >>= 1
def propagates(*ids):
for i in reversed(ids):
v = lazy[i-1]
if v is None:
continue
lazy[2*i-1] = data[2*i-1] = lazy[2*i] = data[2*i] = v
lazy[i-1] = None
def update(l, r, x):
*ids, = gindex(l, r)
propagates(*ids)
L = N0 + l; R = N0 + r
while L < R:
if R & 1:
R -= 1
lazy[R-1] = data[R-1] = x
if L & 1:
lazy[L-1] = data[L-1] = x
L += 1
L >>= 1; R >>= 1
for i in ids:
data[i-1] = min(data[2*i-1], data[2*i])
def query(l, r):
propagates(*gindex(l, r))
L = N0 + l; R = N0 + r
s = INF
while L < R:
if R & 1:
R -= 1
s = min(s, data[R-1])
if L & 1:
s = min(s, data[L-1])
L += 1
L >>= 1; R >>= 1
return s
ans = []
for q in range(Q):
t, *cmd = map(int, readline().split())
if t:
s, t = cmd
ans.append(str(query(s, t+1)))
else:
s, t, x = cmd
update(s, t+1, x)
write("\n".join(ans))
write("\n")
``` | output | 1 | 100,305 | 5 | 200,611 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(s, t): report the minimum element in as, as+1, ..., at.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 s t
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(s, t).
Output
For each find operation, print the minimum value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0 2
1 1 2
Output
1
2
Input
1 3
1 0 0
0 0 0 5
1 0 0
Output
2147483647
5 | instruction | 0 | 100,306 | 5 | 200,612 |
"Correct Solution:
```
import sys
input=sys.stdin.readline
def gindex(l,r):
L,R=l+N0,r+N0
lm=L//(L&-L)>>1
rm=R//(R&-R)>>1
while L<R:
if R<=rm:
yield R
if L<=lm:
yield L
L>>=1
R>>=1
while L:
yield L
L>>=1
def propagates(*ids):
for i in reversed(ids):
v=lazy[i-1]
if v==None:
continue
lazy[2*i-1]=v
lazy[2*i]=v
data[2*i-1]=v
data[2*i]=v
lazy[i-1]=None
def update(l,r,x):#1-index [s,t)
L,R=N0+l,N0+r
*ids,=gindex(l,r)
propagates(*ids)
while L<R:#上から
if R&1:
R-=1
lazy[R-1]=x
data[R-1]=x
if L&1:
lazy[L-1]=x
data[L-1]=x
L+=1
L>>=1
R>>=1
for i in ids:#下から
data[i-1]=min(data[2*i-1],data[2*i])
def query(l,r):
propagates(*gindex(l,r))
L,R=N0+l,N0+r
s=INF
while L<R:
if R&1:
R-=1
s=min(s,data[R-1])
if L&1:
s=min(s,data[L-1])
L+=1
L>>=1
R>>=1
return s
n,q=map(int,input().split())
LV=n.bit_length()
N0=2**LV
INF=2**40
data=[2**31-1]*2*N0
lazy=[None]*2*N0
for i in range(q):
l=list(map(int,input().split()))
if l[0]==0:
s,t,x=l[1:]
update(s,t+1,x)
else:
s,t=l[1:]
print(query(s,t+1))
``` | output | 1 | 100,306 | 5 | 200,613 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(s, t): report the minimum element in as, as+1, ..., at.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 s t
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(s, t).
Output
For each find operation, print the minimum value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0 2
1 1 2
Output
1
2
Input
1 3
1 0 0
0 0 0 5
1 0 0
Output
2147483647
5 | instruction | 0 | 100,307 | 5 | 200,614 |
"Correct Solution:
```
import math
from collections import deque
class SegmentTree:
__slots__ = ["rank", "elem_size", "tree_size", "tree", "lazy", "default_value"]
def __init__(self, a: list, default: int):
self.default_value = default
real_size = len(a)
self.rank = math.ceil(math.log2(real_size))
self.elem_size = 1 << self.rank
self.tree_size = 2 * self.elem_size
self.tree = [default]*self.elem_size + a + [default]*(self.elem_size - real_size)
self.lazy = [None]*self.tree_size
self.init_tree()
def init_tree(self):
tree = self.tree
for i in range(self.elem_size-1, 0, -1):
left, right = tree[i << 1], tree[(i << 1)+1]
# ===== change me =====
tree[i] = left if left < right else right
def propagate(self, l: int, r: int, value: int = None):
'''[x, y)'''
tree, lazy, elem_size, rank = self.tree, self.lazy, self.elem_size, self.rank-1
l, r, targets, p_l, p_r, l_rank, r_rank = l+elem_size, r+elem_size, deque(), 0, 0, 0, 0
t_ap = targets.append
while l < r:
if l & 1:
t_ap(l)
p_l = p_l or l >> 1
l_rank = l_rank or rank
l += 1
if r & 1:
r -= 1
t_ap(r)
p_r = p_r or r >> 1
r_rank = r_rank or rank
l >>= 1
r >>= 1
rank -= 1
deepest = (p_l, p_r)
paths = [[p_l >> n for n in range(l_rank-1, -1, -1)], [p_r >> n for n in range(r_rank-1, -1, -1)]]
for a in paths:
for i in a:
if lazy[i] is None:
continue
# ===== change me =====
tree[i] = lazy[i]
if i < elem_size:
lazy[i << 1] = lazy[i]
lazy[(i << 1)+1] = lazy[i]
lazy[i] = None
result = self.default_value
if value is None:
for i in targets:
if lazy[i] is not None:
tree[i] = lazy[i]
if i < elem_size:
lazy[i << 1] = lazy[i]
lazy[(i << 1)+1] = lazy[i]
lazy[i] = None
# ===== change me =====
if result > tree[i]:
result = tree[i]
else:
for i in targets:
# ===== change me =====
if i < elem_size:
lazy[i << 1] = value
lazy[(i << 1)+1] = value
tree[i] = value
lazy[i] = None
self.update_tree(deepest)
return result
def update_lazy(self, l, r, value):
self.propagate(l, r, value)
def get_value(self, l: int, r: int):
tree = self.tree
targets = self.propagate(l, r)
# ===== change me =====
return min(tree[n] for n in targets)
def update_tree(self, indexes: tuple):
''' ????????????lazy?????¨????????¬????????§???????????¨???????????¨???????????? '''
tree, lazy = self.tree, self.lazy
for k in indexes:
while k > 0:
left, right = k << 1, (k << 1)+1
# ===== change me =====
l_value = tree[left] if lazy[left] is None else lazy[left]
r_value = tree[right] if lazy[right] is None else lazy[right]
tree[k] = l_value if l_value < r_value else r_value
k >>= 1
n, q = map(int, input().split())
rmq = SegmentTree([2**31-1]*n, 2**31-1)
ans = []
append = ans.append
for _ in [0]*q:
l = list(map(int, input().split()))
if l[0] == 0:
rmq.propagate(l[1], l[2]+1, l[3])
else:
a = rmq.propagate(l[1], l[2]+1)
append(a)
print("\n".join([str(n) for n in ans]))
``` | output | 1 | 100,307 | 5 | 200,615 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(s, t): report the minimum element in as, as+1, ..., at.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 s t
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(s, t).
Output
For each find operation, print the minimum value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0 2
1 1 2
Output
1
2
Input
1 3
1 0 0
0 0 0 5
1 0 0
Output
2147483647
5 | instruction | 0 | 100,308 | 5 | 200,616 |
"Correct Solution:
```
import sys
readline = sys.stdin.readline
write = sys.stdout.write
N, Q = map(int, input().split())
INF = 2**31-1
LV = (N-1).bit_length()
N0 = 2**LV
data = [INF]*(2*N0)
lazy = [None]*(2*N0)
def gindex(l, r):
L = (l + N0) >> 1; R = (r + N0) >> 1
lc = 0 if l & 1 else (L & -L).bit_length()
rc = 0 if r & 1 else (R & -R).bit_length()
for i in range(LV):
if rc <= i:
yield R
if L < R and lc <= i:
yield L
L >>= 1; R >>= 1
def propagates(*ids):
for i in reversed(ids):
v = lazy[i-1]
if v is None:
continue
lazy[2*i-1] = data[2*i-1] = lazy[2*i] = data[2*i] = v
lazy[i-1] = None
def update(l, r, x):
*ids, = gindex(l, r)
propagates(*ids)
L = N0 + l; R = N0 + r
while L < R:
if R & 1:
R -= 1
lazy[R-1] = data[R-1] = x
if L & 1:
lazy[L-1] = data[L-1] = x
L += 1
L >>= 1; R >>= 1
for i in ids:
data[i-1] = min(data[2*i-1], data[2*i])
def query(l, r):
propagates(*gindex(l, r))
L = N0 + l; R = N0 + r
s = INF
while L < R:
if R & 1:
R -= 1
s = min(s, data[R-1])
if L & 1:
s = min(s, data[L-1])
L += 1
L >>= 1; R >>= 1
return s
answer = []
for q in range(Q):
t, *cmd = map(int, readline().split())
if t:
s, t = cmd
answer.append(str(query(s, t+1)))
else:
s, t, x = cmd
update(s, t+1, x)
write("\n".join(answer))
write("\n")
``` | output | 1 | 100,308 | 5 | 200,617 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(s, t): report the minimum element in as, as+1, ..., at.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 s t
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(s, t).
Output
For each find operation, print the minimum value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0 2
1 1 2
Output
1
2
Input
1 3
1 0 0
0 0 0 5
1 0 0
Output
2147483647
5 | instruction | 0 | 100,309 | 5 | 200,618 |
"Correct Solution:
```
import sys
input = sys.stdin.readline
N,Q=map(int,input().split())
seg_el=1<<(N.bit_length()) # Segment treeの台の要素数
SEG=[(1<<31)-1]*(2*seg_el) # 1-indexedなので、要素数2*seg_el.Segment treeの初期値で初期化
LAZY=[None]*(2*seg_el) # 1-indexedなので、要素数2*seg_el.Segment treeの初期値で初期化
def indexes(L,R):
INDLIST=[]
R-=1
L>>=1
R>>=1
while L!=R:
if L>R:
INDLIST.append(L)
L>>=1
else:
INDLIST.append(R)
R>>=1
while L!=0:
INDLIST.append(L)
L>>=1
return INDLIST
def updates(l,r,x): # 区間[l,r)をxに更新
L=l+seg_el
R=r+seg_el
L//=(L & (-L))
R//=(R & (-R))
UPIND=indexes(L,R)
for ind in UPIND[::-1]:
if LAZY[ind]!=None:
LAZY[ind<<1]=LAZY[1+(ind<<1)]=SEG[ind<<1]=SEG[1+(ind<<1)]=LAZY[ind]
LAZY[ind]=None
while L!=R:
if L > R:
SEG[L]=x
LAZY[L]=x
L+=1
L//=(L & (-L))
else:
R-=1
SEG[R]=x
LAZY[R]=x
R//=(R & (-R))
for ind in UPIND:
SEG[ind]=min(SEG[ind<<1],SEG[1+(ind<<1)])
def getvalues(l,r): # 区間[l,r)に関するminを調べる
L=l+seg_el
R=r+seg_el
L//=(L & (-L))
R//=(R & (-R))
UPIND=indexes(L,R)
for ind in UPIND[::-1]:
if LAZY[ind]!=None:
LAZY[ind<<1]=LAZY[1+(ind<<1)]=SEG[ind<<1]=SEG[1+(ind<<1)]=LAZY[ind]
LAZY[ind]=None
ANS=1<<31
while L!=R:
if L > R:
ANS=min(ANS , SEG[L])
L+=1
L//=(L & (-L))
else:
R-=1
ANS=min(ANS , SEG[R])
R//=(R & (-R))
return ANS
ANS=[]
for _ in range(Q):
query=list(map(int,input().split()))
if query[0]==0:
updates(query[1],query[2]+1,query[3])
else:
ANS.append(getvalues(query[1],query[2]+1))
print("\n".join([str(ans) for ans in ANS]))
``` | output | 1 | 100,309 | 5 | 200,619 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(s, t): report the minimum element in as, as+1, ..., at.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 s t
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(s, t).
Output
For each find operation, print the minimum value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0 2
1 1 2
Output
1
2
Input
1 3
1 0 0
0 0 0 5
1 0 0
Output
2147483647
5 | instruction | 0 | 100,310 | 5 | 200,620 |
"Correct Solution:
```
# -*- coding: utf-8 -*-
import sys
def input(): return sys.stdin.readline().strip()
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
def list4d(a, b, c, d, e): return [[[[e] * d for j in range(c)] for j in range(b)] for i in range(a)]
def ceil(x, y=1): return int(-(-x // y))
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(N=None): return list(MAP()) if N is None else [INT() for i in range(N)]
def Yes(): print('Yes')
def No(): print('No')
def YES(): print('YES')
def NO(): print('NO')
sys.setrecursionlimit(10 ** 9)
INF = 2 ** 31 - 1
MOD = 10 ** 9 + 7
class SegTreeLazy:
""" 遅延評価セグメント木 """
def __init__(self, N, func, intv):
self.intv = intv
self.func = func
LV = (N-1).bit_length()
self.N0 = 2**LV
self.data = [intv]*(2*self.N0)
self.lazy = [None]*(2*self.N0)
# 伝搬される区間のインデックス(1-indexed)を全て列挙するgenerator
def gindex(self, l, r):
L = l + self.N0; R = r + self.N0
lm = (L // (L & -L)) >> 1
rm = (R // (R & -R)) >> 1
while L < R:
if R <= rm:
yield R
if L <= lm:
yield L
L >>= 1; R >>= 1
while L:
yield L
L >>= 1
# 遅延評価の伝搬処理
def propagates(self, *ids):
# 1-indexedで単調増加のインデックスリスト
for i in reversed(ids):
v = self.lazy[i-1]
if v is None:
continue
self.lazy[2*i-1] = self.data[2*i-1] = self.lazy[2*i] = self.data[2*i] = v
self.lazy[i-1] = None
def update(self, l, r, x):
""" 区間[l,r)の値をxに更新 """
*ids, = self.gindex(l, r)
# 1. トップダウンにlazyの値を伝搬
self.propagates(*ids)
# 2. 区間[l,r)のdata, lazyの値を更新
L = self.N0 + l; R = self.N0 + r
while L < R:
if R & 1:
R -= 1
self.lazy[R-1] = self.data[R-1] = x
if L & 1:
self.lazy[L-1] = self.data[L-1] = x
L += 1
L >>= 1; R >>= 1
# 3. 伝搬させた区間について、ボトムアップにdataの値を伝搬する
for i in ids:
self.data[i-1] = self.func(self.data[2*i-1], self.data[2*i])
def query(self, l, r):
""" 区間[l,r)の最小値を取得 """
# 1. トップダウンにlazyの値を伝搬
self.propagates(*self.gindex(l, r))
L = self.N0 + l; R = self.N0 + r
# 2. 区間[l, r)の最小値を求める
s = self.intv
while L < R:
if R & 1:
R -= 1
s = self.func(s, self.data[R-1])
if L & 1:
s = self.func(s, self.data[L-1])
L += 1
L >>= 1; R >>= 1
return s
N, Q = MAP()
stl = SegTreeLazy(N+1, min, INF)
ans = []
for i in range(Q):
cmd, *arg = MAP()
if cmd == 0:
s, t, x = arg
stl.update(s, t+1, x)
else:
s, t = arg
ans.append(str(stl.query(s, t+1)))
print('\n'.join(ans))
``` | output | 1 | 100,310 | 5 | 200,621 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(s, t): report the minimum element in as, as+1, ..., at.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 s t
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(s, t).
Output
For each find operation, print the minimum value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0 2
1 1 2
Output
1
2
Input
1 3
1 0 0
0 0 0 5
1 0 0
Output
2147483647
5
Submitted Solution:
```
import sys
readline = sys.stdin.readline
write = sys.stdout.write
N, Q = map(int, input().split())
INF = 2**31-1
LV = (N-1).bit_length()
N0 = 2**LV
data = [INF]*(2*N0)
lazy = [None]*(2*N0)
def gindex(l, r):
L = (l + N0) >> 1; R = (r + N0) >> 1
lc = 0 if l & 1 else (L & -L).bit_length()
rc = 0 if r & 1 else (R & -R).bit_length()
for i in range(LV):
if rc <= i:
yield R
if L < R and lc <= i:
yield L
L >>= 1; R >>= 1
def propagates(*ids):
for i in reversed(ids):
v = lazy[i-1]
if v is None:
continue
lazy[2*i-1] = data[2*i-1] = lazy[2*i] = data[2*i] = v
lazy[i-1] = None
def update(l, r, x):
*ids, = gindex(l, r)
propagates(*ids)
L = N0 + l; R = N0 + r
while L < R:
if R & 1:
R -= 1
lazy[R-1] = data[R-1] = x
if L & 1:
lazy[L-1] = data[L-1] = x
L += 1
L >>= 1; R >>= 1
for i in ids:
data[i-1] = min(data[2*i-1], data[2*i])
def query(l, r):
propagates(*gindex(l, r))
L = N0 + l; R = N0 + r
s = INF
while L < R:
if R & 1:
R -= 1
s = min(s, data[R-1])
if L & 1:
s = min(s, data[L-1])
L += 1
L >>= 1; R >>= 1
return s
ans = []
for q in range(Q):
t, *cmd = map(int, readline().split())
if t:
s, t = cmd
ans.append(str(query(s, t+1)))
else:
s, t, x = cmd
update(s, t+1, x)
write("\n".join(ans))
write("\n")
``` | instruction | 0 | 100,311 | 5 | 200,622 |
Yes | output | 1 | 100,311 | 5 | 200,623 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(s, t): report the minimum element in as, as+1, ..., at.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 s t
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(s, t).
Output
For each find operation, print the minimum value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0 2
1 1 2
Output
1
2
Input
1 3
1 0 0
0 0 0 5
1 0 0
Output
2147483647
5
Submitted Solution:
```
class LazySegTree: # Non Recursion, RmQ, RUQ
def __init__(self, N):
self.N = N
self._N = 1<<((N-1).bit_length())
self.INF = (1<<31) - 1
self.node = [self.INF]*2*self._N
self.lazy = [None]*2*self._N
def build(self, lis):
for i in range(self.N):
self.node[i+self._N-1] = lis[i]
for i in range(self._N-2,-1,-1):
self.node[i] = self.node[i*2+1] + self.node[i*2+2]
def _gindex(self, l, r):
left = l + self._N; right = r + self._N
lm = (left // (left & -left)) >> 1
rm = (right // (right & -right)) >> 1
while left < right:
if right <= rm: yield right
if left <= lm: yield left
left >>= 1; right >>= 1
while left:
yield left
left >>= 1
def propagates(self, *ids): # ids: 1-indexded
for i in reversed(ids):
v = self.lazy[i-1]
if v is None: continue
self.lazy[2*i-1] = self.node[2*i-1] = v
self.lazy[2*i] = self.node[2*i] = v
self.lazy[i-1] = None
return
def update(self, l, r, x): # change all[left, right) to x
*ids, = self._gindex(l, r)
self.propagates(*ids)
left = l + self._N; right = r + self._N
while left < right:
if left & 1:
self.lazy[left-1] = self.node[left-1] = x
left += 1
if right & 1:
right -= 1
self.lazy[right-1] = self.node[right-1] = x
left >>= 1; right >>= 1
for i in ids:
self.node[i-1] = min(self.node[2*i-1], self.node[2*i])
def query(self, l, r):
self.propagates(*self._gindex(l, r))
left = l + self._N; right = r + self._N
ret = self.INF
while left < right:
if right & 1:
right -= 1
ret = min(ret, self.node[right-1])
if left & 1:
ret = min(ret, self.node[left-1])
left += 1
left >>= 1; right >>= 1
return ret
nm = lambda: map(int, input().split())
n,q = nm()
S = LazySegTree(n)
for _ in range(q):
v = list(nm())
if v[0]:
# print(S.node)
# print(S.lazy)
print(S.query(v[1], v[2]+1))
else:
S.update(v[1], v[2]+1, v[3])
``` | instruction | 0 | 100,312 | 5 | 200,624 |
Yes | output | 1 | 100,312 | 5 | 200,625 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(s, t): report the minimum element in as, as+1, ..., at.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 s t
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(s, t).
Output
For each find operation, print the minimum value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0 2
1 1 2
Output
1
2
Input
1 3
1 0 0
0 0 0 5
1 0 0
Output
2147483647
5
Submitted Solution:
```
class LazyPropSegmentTree:
def __init__(self, lst, op, apply, comp, e, identity):
self.n = len(lst)
self.depth = (self.n - 1).bit_length()
self.N = 1 << self.depth
self.op = op # binary operation of elements
self.apply = apply # function to apply to an element
self.comp = comp # composition of functions
self.e = e # identity element w.r.t. op
self.identity = identity # identity element w.r.t. comp
self.v = self._build(lst) # self.v is set to be 1-indexed for simplicity
self.lazy = [self.identity] * (2 * self.N)
def __getitem__(self, i):
return self.fold(i, i+1)
def _build(self, lst):
# construction of a tree
# total 2 * self.N elements (tree[0] is not used)
tree = [self.e] * (self.N) + lst + [self.e] * (self.N - self.n)
for i in range(self.N - 1, 0, -1): tree[i] = self.op(tree[i << 1], tree[(i << 1)|1])
return tree
def _indices(self, l, r):
left = l + self.N; right = r + self.N
left //= (left & (-left)); right //= (right & (-right))
left >>= 1; right >>= 1
while left != right:
if left > right: yield left; left >>= 1
else: yield right; right >>= 1
while left > 0: yield left; left >>= 1
# propagate self.lazy and self.v in a top-down manner
def _propagate_topdown(self, *indices):
identity, v, lazy, apply, comp = self.identity, self.v, self.lazy, self.apply, self.comp
for k in reversed(indices):
x = lazy[k]
if x == identity: continue
lazy[k << 1] = comp(lazy[k << 1], x)
lazy[(k << 1)|1] = comp(lazy[(k << 1)|1], x)
v[k << 1] = apply(v[k << 1], x)
v[(k << 1)|1] = apply(v[(k << 1)|1], x)
lazy[k] = identity # propagated
# propagate self.v in a bottom-up manner
def _propagate_bottomup(self, indices):
v, op = self.v, self.op
for k in indices: v[k] = op(v[k << 1], v[(k << 1)|1])
# update for the query interval [l, r) with function x
def update(self, l, r, x):
*indices, = self._indices(l, r)
self._propagate_topdown(*indices)
N, v, lazy, apply, comp = self.N, self.v, self.lazy, self.apply, self.comp
# update self.v and self.lazy for the query interval [l, r)
left = l + N; right = r + N
if left & 1: v[left] = apply(v[left], x); left += 1
if right & 1: right -= 1; v[right] = apply(v[right], x)
left >>= 1; right >>= 1
while left < right:
if left & 1:
lazy[left] = comp(lazy[left], x)
v[left] = apply(v[left], x)
left += 1
if right & 1:
right -= 1
lazy[right] = comp(lazy[right], x)
v[right] = apply(v[right], x)
left >>= 1; right >>= 1
self._propagate_bottomup(indices)
# returns answer for the query interval [l, r)
def fold(self, l, r):
self._propagate_topdown(*self._indices(l, r))
e, N, v, op = self.e, self.N, self.v, self.op
# calculate the answer for the query interval [l, r)
left = l + N; right = r + N
L = R = e
while left < right:
if left & 1: # self.v[left] is the right child
L = op(L, v[left])
left += 1
if right & 1: # self.v[right-1] is the left child
right -= 1
R = op(v[right], R)
left >>= 1; right >>= 1
return op(L, R)
N, Q = map(int, input().split())
op = min
apply = lambda x, f: f
comp = lambda f, g: g
e = 2**31 - 1
identity = None
A = [e] * N
lpsg = LazyPropSegmentTree(A, op, apply, comp, e, identity)
ans = []
for _ in range(Q):
t, *arg, = map(int, input().split())
if t == 0:
s, t, x = arg
lpsg.update(s, t+1, x)
else:
s, t = arg
ans.append(lpsg.fold(s, t+1))
print('\n'.join(map(str, ans)))
``` | instruction | 0 | 100,313 | 5 | 200,626 |
Yes | output | 1 | 100,313 | 5 | 200,627 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(s, t): report the minimum element in as, as+1, ..., at.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 s t
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(s, t).
Output
For each find operation, print the minimum value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0 2
1 1 2
Output
1
2
Input
1 3
1 0 0
0 0 0 5
1 0 0
Output
2147483647
5
Submitted Solution:
```
import sys
input = sys.stdin.readline
n, q = map(int, input().split())
INF = 2**31-1
Len = (n-1).bit_length()
size = 2**Len
tree = [INF]*(2*size)
lazy = [None]*(2*size)
def gindex(l,r):
L = (l+size)>>1;R=(r+size)>>1
lc = 0 if l & 1 else (L&-L).bit_length()
rc = 0 if r & 1 else (R&-R).bit_length()
for i in range(Len):
if rc <= i:
yield R
if L < R and lc <= i:
yield L
L >>= 1; R >>= 1
def propagates(ids):
#上から更新
for i in reversed(ids):
v = lazy[i]
if v is None:
continue
lazy[2*i]=tree[2*i]=lazy[2*i+1]=tree[2*i+1]=v
lazy[i]=None
#[l,r)の探索
def update(l,r,x):
*ids, = gindex(l, r)
propagates(ids)
L = size+l
R = size+r
while L<R:
if R&1:
R -= 1
lazy[R]=tree[R]=x
if L&1:
lazy[L]=tree[L]=x
L+=1
L>>=1;R>>=1
#値を更新し終わったら、最小値を下から更新
for i in ids:
if 2*i+1<size*2:
tree[i]=min(tree[i*2],tree[i*2+1])
def query(l, r):
*ids, = gindex(l, r)
propagates(ids)
L = size + l
R = size + r
s = INF
while L<R:
if R&1:
R-=1
s = min(s,tree[R])
if L&1:
s = min(s,tree[L])
L+=1
L>>=1;R>>=1
return s
ans = []
for i in range(q):
a, *b = map(int, input().split())
if a:
ans.append(query(b[0],b[1]+1))
else:
update(b[0],b[1]+1,b[2])
print('\n'.join(map(str,ans)))
``` | instruction | 0 | 100,314 | 5 | 200,628 |
Yes | output | 1 | 100,314 | 5 | 200,629 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(s, t): report the minimum element in as, as+1, ..., at.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 s t
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(s, t).
Output
For each find operation, print the minimum value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0 2
1 1 2
Output
1
2
Input
1 3
1 0 0
0 0 0 5
1 0 0
Output
2147483647
5
Submitted Solution:
```
import math
lazy_default = None
def update_tree(k: int):
if k >= l:
k >>= l
while k > 0:
left, right = tree[k << 1], tree[(k << 1) + 1]
# ===== change me =====
tree[k] = left if left < right else right
k >>= 1
def merge(node_index, value, update=False):
lazy_value = lazy[node_index]
# ===== change me =====
value = lazy_value if value is None else value
lazy[node_index] = None
# ===== change me =====
if value is not lazy_default:
tree[node_index] = value
if node_index < l and value != lazy_default:
# ===== change me =====
if node_index < l // 2:
tree[node_index << 1] = value
tree[(node_index << 1) + 1] = value
else:
lazy[node_index << 1] = value
lazy[(node_index << 1) + 1] = value
def hoge(s, e, value):
s, e, d, l_active, r_active = s+l, e+l, rank, 1, 1
values = []
append = values.append
while l_active or r_active:
l_node, r_node = s >> d, e >> d
l_leaf = l_node << d, (l_node + 1 << d) - 1
r_leaf = r_node << d, (r_node + 1 << d) - 1
lazy_left, lazy_right = lazy[l_node], lazy[r_node]
if l_node == r_node:
if l_leaf[0] == s and r_leaf[1] == e:
# match
merge(l_node, value, True)
append(tree[l_node])
break
else:
if l_active:
if l_leaf[0] == s:
# match
merge(l_node, value, True)
append(tree[l_node])
l_active = 0
# ??????????????????????????????????????????????¶?????????????????????´???
else:
if lazy_left != None:
merge(l_node, None)
if l_node >> 1 << 1 == l_node and l_node+1 < r_node:
# match
merge(l_node+1, value)
append(tree[l_node+1])
if r_active:
if r_leaf[1] == e:
# match
merge(r_node, value)
update_tree(r_node)
append(tree[r_node])
r_active = 0
else:
if lazy_right != None:
merge(r_node, None)
if r_node >> 1 << 1 == r_node-1 and l_node+1 < r_node:
# match
merge(r_node-1, value)
append(tree[r_node-1])
d -= 1
return values
def get_value(i):
i += l
v = 0
for j in range(rank, -1, -1):
v += lazy[i>>j]
return v + tree[i]
n, q = map(int,input().split())
l = 1 << math.ceil(math.log2(n))
tree = [2**31-1]*(2*l)
lazy = [None]*(2*l)
rank = int(math.log2(l))
ans = []
ap = ans.append
for _ in [None]*q:
query = list(map(int, input().split()))
if query[0] == 0:
hoge(query[1], query[2], query[3])
else:
a = hoge(query[1], query[2], None)
ap(min(a))
print("\n".join((str(n) for n in ans)))
``` | instruction | 0 | 100,315 | 5 | 200,630 |
No | output | 1 | 100,315 | 5 | 200,631 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(s, t): report the minimum element in as, as+1, ..., at.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 s t
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(s, t).
Output
For each find operation, print the minimum value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0 2
1 1 2
Output
1
2
Input
1 3
1 0 0
0 0 0 5
1 0 0
Output
2147483647
5
Submitted Solution:
```
from math import log2, ceil
class SegmentTree:
SENTINEL = 2 ** 31 - 1
def __init__(self, n):
tn = 2 ** ceil(log2(n))
self.a = [2 ** 31 - 1] * (tn * 2)
def find(self, c, l, r, s, t):
if l == s and r == t:
return self.a[c]
mid = (l + r) // 2
if t <= mid:
return self.find(c * 2, l, mid, s, t)
elif s > mid:
return self.find(c * 2 + 1, mid + 1, r, s, t)
else:
return min(
self.find(c * 2, l, mid, s, mid),
self.find(c * 2 + 1, mid + 1, r, mid + 1, t))
def update(self, c, l, r, s, t, x):
if l == s and r == t:
self.a[c] = x
return x
mid = (l + r) // 2
if t <= mid:
if self.a[c * 2 + 1] == self.SENTINEL:
self.a[c * 2 + 1] = self.a[c]
u = min(self.update(c * 2, l, mid, s, t, x), self.a[c * 2 + 1])
elif s > mid:
if self.a[c * 2] == self.SENTINEL:
self.a[c * 2] = self.a[c]
u = min(self.a[c * 2], self.update(c * 2 + 1, mid + 1, r, s, t, x))
else:
u = min(
self.update(c * 2, l, mid, s, mid, x),
self.update(c * 2 + 1, mid + 1, r, mid + 1, t, x))
self.a[c] = u
return u
n, q = map(int, input().split())
st = SegmentTree(n)
for _ in range(q):
query = input().split()
if query[0] == '0':
s, t, x = map(int, query[1:])
st.update(1, 0, n - 1, s, t, x)
else:
s, t = map(int, query[1:])
print(st.find(1, 0, n - 1, s, t))
``` | instruction | 0 | 100,316 | 5 | 200,632 |
No | output | 1 | 100,316 | 5 | 200,633 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(s, t): report the minimum element in as, as+1, ..., at.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 s t
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(s, t).
Output
For each find operation, print the minimum value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0 2
1 1 2
Output
1
2
Input
1 3
1 0 0
0 0 0 5
1 0 0
Output
2147483647
5
Submitted Solution:
```
INT_MAX = 1<<31 - 1
_N,query_num = map(int, input().split())
queries = [list(map(int, input().split())) for i in range(query_num)]
N = 1<<17
arr_size = 2*N - 1
A = [INT_MAX]*(arr_size)
lazy_A = [None]*(arr_size)
def tree_propagate(k):
if(lazy_A[k] is None):
return
if(k < N // 2):
A[k*2+1] = A[k*2+2] = lazy_A[k*2+1] = lazy_A[k*2+2] = lazy_A[k]
lazy_A[k] = None
return
if(k < N-1):
A[k*2+1] = A[k*2+2] = lazy_A[k]
lazy_A[k] = None
def update(s,t,x,k,l,r):
if (r <= s or t <= l):
return
if (s <= l and r <= t):
A[k] = x
if(k < N-1):
lazy_A[k] = x
return
tree_propagate(k)
center = (l+r)//2
update(s, t, x, k*2 + 1, l, center)
update(s, t, x, k*2 + 2, center, r)
A[k] = min(A[k*2+1], A[k*2+2])
def find(s,t,k,l,r):
if (r <= s or t <= l):
return INT_MAX
if(s <= l and r <= t):
return A[k]
tree_propagate(k)
center = (l+r)//2
v_left = find(s, t, k*2 + 1, l, center)
v_right = find(s, t, k*2 + 2, center, r)
return min(v_left,v_right)
answers = []
for query in queries:
# update
if(query[0]==0):
update(query[1],query[2]+1,query[3],0,0,N)
# find
else:
answers.append(find(query[1],query[2]+1,0,0,N))
print("\n".join([str(ans) for ans in answers]))
``` | instruction | 0 | 100,317 | 5 | 200,634 |
No | output | 1 | 100,317 | 5 | 200,635 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program which manipulates a sequence A = {a0, a1, . . . , an−1} with the following operations:
* update(s, t, x): change as, as+1, ..., at to x.
* find(s, t): report the minimum element in as, as+1, ..., at.
Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 0 ≤ s ≤ t < n
* 0 ≤ x < 231−1
Input
n q
query1
query2
:
queryq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:
0 s t x
or
1 s t
The first digit represents the type of the query. '0' denotes update(s, t, x) and '1' denotes find(s, t).
Output
For each find operation, print the minimum value.
Examples
Input
3 5
0 0 1 1
0 1 2 3
0 2 2 2
1 0 2
1 1 2
Output
1
2
Input
1 3
1 0 0
0 0 0 5
1 0 0
Output
2147483647
5
Submitted Solution:
```
import math
class SegmentTree:
__slots__ = ["elem_size", "tree_size", "tree", "lazy"]
def __init__(self, a: list, default: int):
real_size = len(a)
self.elem_size = 1 << math.ceil(math.log2(real_size))
self.tree_size = 2 * self.elem_size
self.tree = [default]*self.elem_size + a + [default]*(self.elem_size - real_size)
self.lazy = [None]*self.tree_size
self.init_tree()
def init_tree(self):
tree = self.tree
for i in range(self.elem_size-1, 0, -1):
left, right = tree[i << 1], tree[(i << 1)+1]
# ===== change me =====
tree[i] = left if left < right else right
def get_indexes(self, l: int, r: int):
'''[x, y)'''
l, r, indexes = l+self.elem_size, r+self.elem_size, []
append = indexes.append
while l < r:
if l & 1:
append(l)
l += 1
if r & 1:
r -= 1
append(r)
l, r = l >> 1, r >> 1
return indexes
def get_indexes_with_propagation(self, l: int, r: int, current_node: int, l_end: int, r_end: int):
# print(l,r,current_node,l_end,r_end,self.lazy[current_node])
indexes = []
tree, lazy = self.tree, self.lazy
lazy_value = lazy[current_node]
left_child, right_child = current_node << 1, (current_node << 1) + 1
if lazy_value is not None:
self.lazy[current_node] = None
tree[current_node] = lazy_value
if left_child < self.tree_size:
if left_child < self.elem_size:
lazy[left_child] = lazy[right_child] = lazy_value
else:
tree[left_child] = tree[right_child] = lazy_value
if l == l_end and r == r_end:
return [current_node]
mid = (l_end + r_end) // 2
if l < mid:
l_r = r if r < mid else mid
indexes += self.get_indexes_with_propagation(l, l_r, left_child, l_end, mid)
if r > mid:
r_l = l if mid < l else mid
indexes += self.get_indexes_with_propagation(r_l, r, right_child, mid, r_end)
return indexes
def update_lazy(self, l, r, value):
indexes = sorted(self.get_indexes_with_propagation(l, r, 1, 0, self.elem_size))
l = self.elem_size
lazy, tree = self.lazy, self.tree
for n in indexes:
lazy[n] = None
if n < l//2:
lazy[n << 1] = value
lazy[(n << 1)+1] = value
elif n < l:
tree[n << 1] = value
tree[(n << 1)+1] = value
else:
tree[n] = value
self.update_tree(n)
def get_value(self, l: int, r: int):
tree = self.tree
index_list = self.get_indexes_with_propagation(l, r, 1, 0, self.elem_size)
# ===== change me =====
return min(tree[n] for n in index_list)
def update_tree(self, k: int):
tree, lazy = self.tree, self.lazy
if k >= self.elem_size:
k >>= 1
while k > 1:
left, right = k << 1, (k << 1)+1
left_value = lazy[left] or tree[left]
right_value = lazy[right] or tree[right]
# ===== change me =====
tree[k] = left_value if left_value < right_value else right_value
k >>= 1
def set_value(self, i: int, value: int, op: str):
k = self.elem_size + i
if op == "=":
self.tree[k] = value
elif op == "+":
self.tree[k] += value
self.update_tree(k)
n, q = map(int, input().split())
rmq = SegmentTree([2**31-1]*n, 2**31-1)
ans = []
for _ in [0]*q:
l = list(map(int, input().split()))
if l[0] == 0:
rmq.update_lazy(l[1], l[2]+1, l[3])
else:
ans.append(rmq.get_value(l[1], l[2]+1))
print("\n".join([str(n) for n in ans]))
``` | instruction | 0 | 100,318 | 5 | 200,636 |
No | output | 1 | 100,318 | 5 | 200,637 |
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